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Note, first, that

i2 = -1; also, that

(1 + i)2 = 1 + 2i - 1 = 2i; and, finally, that

[±½(√2) ]2 = ½.

Thus, [±½(√2)(1 + i)]2 = i.

Therefore, [±½ (√2)(i - 1)]2 = -i, and

±½ (√2)(i - 1) = √-i.

* * *

In case calculus is familiar, you may prefer to find the desired roots in this manner:

eiθ = cosθ + i sinθ; thus, when

θ = ½ π, eiθ = i; and, when

θ = ¼ π, eiθ = +√i.

Then, +√i = cos ¼ π + i sin ¼ π

= (½√2) + i ½√2

= (½√2) (1 + i).

This gives,

+√-i = (+√i)(+√-1)

= i(+√i)

= (½√2) i(1 + i)

= (½√2) (i - 1).

Therefore the two roots are:

±√-i = ±(½√2)(i - 1).

* * *

The first method is simpler, for proving the result; the second method, for finding it.

Q: How do you find the two square roots of -i?

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